Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question about the irrationality of $e$:

In proving the irrationality of $e$, one can prove the irrationality of $e^{-1}$ by using the series $$e^x = 1+x+\frac{x^2}{2!} + \cdots + \frac{x^n}{n!}+ \cdots$$ So the series for $e^{-1}$ is $$s_n = \sum_{k=0}^{n} \frac{(-1)^{k}}{k!}$$ so that $$0 < e^{-1}-s_{2k-1} < \frac{1}{(2k)!}$$ or $$0 < (2k-1)!(e^{-1}-s_{2k-1}) < \frac{1}{2k} \leq \frac{1}{2}$$ If $e^{-1}$ were rational then we would have a difference of two integers which is not an integer (i.e. between $0$ and $\frac{1}{2}$).

Question. Is this "irrationality of $e^{-1}$ by alternating series" proof what motivated the definition of irrationality measure? It seems that this was borne out of using alternating series.

share|improve this question
4  
I always thought the definition of irrationality measure was motivated from Liouville's theorem: en.wikipedia.org/wiki/… –  Qiaochu Yuan Aug 3 '11 at 5:14
    
I think there is no connection. Incidentally, I prefer going through $e^{-1}$, as you did. Proofs I have seen of the irrationality of $e$ use the series for $e$ directly, which involves some additional work to produce the required estimate. –  André Nicolas Aug 3 '11 at 5:30
2  
This is a really nice proof that I hadn't seen before. Thanks! –  ShreevatsaR Aug 3 '11 at 11:49
    
As far as I know, the "series proof" for the irrationality of $e$ first appeared in the following 1815 book by Janot de Stainville (Article 232, pp. 339-341): books.google.com/books?id=5J0AAAAAMAAJ Stainville says he learned the proof from Poinsot and that the proof itself is due to Fourier. –  Dave L. Renfro Aug 3 '11 at 14:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.